cbse ncert solutions for class 12 physics chapter 8...class- xii-cbse-physics electromagnetic waves...

Class- XII-CBSE-Physics Electromagnetic Waves

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CBSE NCERT Solutions for Class 12 Physics Chapter 8 Back of Chapter Questions

8.1. Figure 8.6 shows a capacitor made of two circular plates, each of radius 12 cm and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of change of the potential differencebetween the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor?Explain.

Solution:

Given that

The magnitude of charging current, I = 0.15 A

Distance between the plates, d = 5 cm = 0.05 m

The Radius of each circular plate,

r = 12 cm = 0.12 m

The permittivity of free space, ∈0 = 8.85 × 10−12 C2 N−1 m−2

(a) We know that the capacitance between the two plates is given by therelation, 𝐶𝐶 = 𝜀𝜀0𝐴𝐴

𝑑𝑑

Where, A = Area of each plate = 𝜋𝜋𝑟𝑟2

C =ε0 × πr2

d=

8.85 × 10−12 × π × (0.12)2

0.05= 8.0032 × 10−12 F

= 80.032 pF

Charge on each plate, 𝑞𝑞 = 𝐶𝐶𝐶𝐶

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Where V = Potential difference across the plates

Differentiate both sides with respect to time (t)

𝑑𝑑𝑞𝑞𝑑𝑑𝑑𝑑

= 𝐶𝐶𝑑𝑑𝐶𝐶𝑑𝑑𝑑𝑑

But, 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= current (𝐼𝐼)

∴𝑑𝑑𝐶𝐶𝑑𝑑𝑑𝑑

=𝐼𝐼𝐶𝐶

⇒0.15

80.032 × 10−12= 1.87 × 109 V s⁄

Therefore, the change in the potential difference between the plates with respect to time is 1.87 × 109 V s⁄

(b) Here the displacement current in the plates is the same as the conduction current Hence, the displacement current, 𝑖𝑖𝑑𝑑 is 0.15 A.

(c) Yes

If we take the sum of conduction and displacement current, then Kirchhoff's first rule is valid at each plate of the capacitor

8.2. A parallel plate capacitor (Figure. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an (angular) frequency of 300 rad s-1

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Solution:

Given that




The capacitance of a parallel plate capacitor, 𝐶𝐶 = 100 pF = 100 × 10−12 F

The radius of each circular plate, 𝑅𝑅 = 6.0 cm = 0.06 m

The capacitance of a parallel plate capacitor, 𝐶𝐶 = 100 pF = 100 × 10−12 F

The voltage supply to the capacitor, 𝐶𝐶 = 230 V

Angular frequency, 𝜔𝜔 = 300 rad s−1

(a) Rms value of conduction current, 𝐼𝐼 = 𝑉𝑉𝑋𝑋𝐶𝐶

Where, 𝑋𝑋𝐶𝐶 = Capacitive reactance = 1𝜔𝜔𝐶𝐶

∴ 𝐼𝐼 = 𝐶𝐶 × 𝜔𝜔𝐶𝐶 = 230 × 300 × 100 × 10−12 = 6.9 × 10−6A = 6.9 μA

Hence, the rms value of the conduction current is 6.9 μA.

(b) Yes, the magnitude of conduction current is equal to displacement current.

(c) The magnetic field is given as 𝐵𝐵 = 𝜇𝜇0𝑟𝑟2𝜋𝜋𝑅𝑅2

𝐼𝐼0

Where, μo = Free space permeability = 4π × 10−7 N A−2

𝐼𝐼𝑜𝑜 = Maximum value of current = √2𝐼𝐼

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

∴ B =4π × 10−7 × 0.03 × √2 × 6.9 × 10−6

2π(0.06)2 = 1.63 × 10−11 T.

Hence, the magnetic field at that point is 1.63 × 10−11 T.

8.3. What physical quantity is the same for X-rays of wavelength 10−10 m red light of wavelength 6800 Å and radio waves of wavelength 500 m?

Solution:

The magnitude of the speed of light (3 × 108 m/s) in a vacuum is the same for all type of wavelengths. It does not depend on the wavelength in the vacuum.

8.4. A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Solution:

Given that

The electromagnetic wave travels along the 𝑧𝑧-direction in a vacuum.

The electric field (E) and the magnetic field (B) are in the 𝑥𝑥 − 𝑦𝑦 plane. They are mutually perpendicular.




Frequency of the wave, 𝑣𝑣 = 30 MHz = 30 × 106 Hz

Speed of light in a vacuum, 𝑐𝑐 = 3 × 108 m/s

We know that expression for the wavelength of a wave is given as:

𝜆𝜆 =𝑐𝑐𝑣𝑣

𝜆𝜆 =3 × 108

30 × 106= 10 m

Hence the wavelength of the electromagnetic wave is 10 m

8.5. A radio can tune in to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?

Solution:

Given that,

The minimum frequency of radio wave, ν1 = 7.5 MHz = 7.5 × 106 Hz

Maximum frequency, 𝜈𝜈2 = 12 MHz = 12 × 106 Hz,

Speed of light, 𝑐𝑐 = 3 × 108 m/s

Wavelength for ν1 can be calculated as:

𝜆𝜆 =𝑐𝑐𝜈𝜈1

=3 × 108

7.5 × 106= 40 m

Wavelength for 𝑣𝑣2 can be calculated as:

𝜆𝜆 =𝑐𝑐𝜈𝜈2

=3 × 108

12 × 106= 25 m

Hence the wavelength range of the radio is 40 m to 25 m.

8.6. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Solution:

About its mean position, the frequency of an electromagnetic wave produced by the oscillator is the same as the frequency of a charged particle oscillating, i.e., 109 Hz.




8.7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0= 510 nT. What is the amplitude of the electric field part of the wave?

Solution:

Given that,

The amplitude of the magnetic field of an electromagnetic wave in a vacuum,

Bo = 510 nT = 510 × 10−9 T

Speed of light in a vacuum, 𝑐𝑐 = 3 × 108 m s⁄

The relation between the amplitude of the electric field and the magnetic field is

𝑐𝑐 = 𝐸𝐸0𝐵𝐵0

𝐸𝐸 = 𝑐𝑐𝐵𝐵𝑜𝑜 = 3 × 108 × 510 × 10−9 = 153 N C⁄

Therefore, the amplitude of the electric field part of the wave is 153 N C⁄ .

8.8. Suppose that the electric field amplitude of an electromagnetic wave is E0=120 N/C and that its frequency is ν = 50.0 MHz.

(a) Determine, B0𝜔𝜔, 𝑘𝑘, and λ.

(b) Find expressions for E and B.

Solution:

Given that,

The amplitude of Electric field, 𝐸𝐸𝑜𝑜 = 120 N/C

Frequency of electromagnetic wave, 𝑣𝑣 = 50.0 MHz = 50 × 106 Hz

Speed of light, 𝑐𝑐 = 3 × 108 m s⁄

(a) The magnitude of magnetic field strength is given as:

𝐵𝐵0 =𝐸𝐸0𝑐𝑐

=120

3 × 108

= 4 × 10−7 T = 400 nT

Angular frequency of the electromagnetic wave is given as:

= 𝜔𝜔 = 2𝜋𝜋𝑣𝑣 = 2𝜋𝜋 × 50 × 106 = 3.14 × 108 rad s⁄

Propagation constant is given as:

𝑘𝑘 =𝜔𝜔𝑐𝑐




=3.14 × 108

3 × 108= 1.05 rad m⁄

The wavelength of the wave is given as:


=3 × 108

50 × 106= 6.0 m

(b) Suppose that the wave is propagating in the positive 𝑥𝑥 direction. Then, the direction of the electric field vector will be in the positive y-direction, and the magnetic field vector will be in the positive 𝑧𝑧 direction because all three vectors are mutually perpendicular.

Equation of electric field vector is given as:

𝐸𝐸�⃗ = 𝐸𝐸0 sin(𝑘𝑘𝑥𝑥 − 𝜔𝜔𝑑𝑑) 𝚥𝚥�

= 120 sin[1.05𝑥𝑥 − 3.14 × 108𝑑𝑑] 𝚥𝚥̂

And, magnetic field vector is given as:

𝐵𝐵�⃗ = 𝐵𝐵0 sin(𝑘𝑘𝑥𝑥 − 𝜔𝜔𝑑𝑑)𝑘𝑘�

𝐵𝐵�⃗ = (4 × 10−7) sin[1.05𝑥𝑥 − 3.14 × 108𝑑𝑑] 𝑘𝑘�

8.9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Solution:

The energy of a photon is given as:

𝐸𝐸 = ℎ𝑣𝑣 =ℎ𝑐𝑐𝜆𝜆

Where,

ℎ = Planck's constant = 6.6 × 10−34 J s

𝑐𝑐 = Speed of light = 3 × 108 m s⁄

𝜆𝜆 = Wavelength of radiation

∴ 𝐸𝐸 =6.6 × 10−34 × 3 × 108

𝜆𝜆=

19.8 × 10−26

𝜆𝜆 J

=19.8 × 10−26

𝜆𝜆 × 1.6 × 10−16=

12.375 × 10−7

𝜆𝜆 eV




The given table lists the photon energies for different parts of an electromagnetic spectrum for different 𝜆𝜆 .

λ (m) 103 1 10−3 10−6 10−8 10−10 10−12

E (eV) 12.37× 10−1

12.375× 10−7

12.375× 10−4

12.375× 10−1

12.375× 101

12.375× 103

12.375× 105

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

8.10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s−1]

Solution:

Given that,

Frequency of the electromagnetic wave, 𝑣𝑣 = 2.0 × 1010 Hz

The amplitude of the electric field, 𝐸𝐸𝑜𝑜 = 48 V m−1

Speed of light, 𝑐𝑐 = 3 × 108 m/s

(a) The wavelength of a wave is given as:


=3 × 108

2 × 1010= 0.015 m

(b) Magnetic field strength is given as:


=48

3 × 108= 1.6 × 10−7 T

(c) The energy density of the electric field is given as:

𝑈𝑈𝐸𝐸 =12𝜀𝜀0 𝐸𝐸2

And, the energy density of the magnetic field is given as:




𝑈𝑈𝐵𝐵 =1

2𝜇𝜇0 𝐵𝐵2

Where,

𝜀𝜀0 = Permittivity of free space

𝜇𝜇0 = Permeability of free space

We have the relation connecting E and B as:

𝐸𝐸 = 𝑐𝑐𝐵𝐵… (1)

Where,

𝑐𝑐 =1

�𝜀𝜀0 𝜇𝜇0… (2)

Putting equation (2) in equation (1), we get

𝐸𝐸 =1

�𝜀𝜀0 𝜇𝜇0 𝐵𝐵

Squaring both sides, we get

𝐸𝐸2 =1

𝜀𝜀0 𝜇𝜇0 𝐵𝐵2

𝜀𝜀0 𝐸𝐸2 =𝐵𝐵2

𝜇𝜇0

12𝜀𝜀0 𝐸𝐸2 =

12𝐵𝐵2

𝜇𝜇0

⇒ 𝑈𝑈𝐸𝐸 = 𝑈𝑈𝐵𝐵

Hence the average energy density of the E field equals the average energy density of the B field

Additional Exercises

8.11. Suppose that the electric field part of an electromagnetic wave in vacuum is

E = {(3.1 N/C)cos[(1.8 rad/m)y + (5.4 × 106rad/s)t]}�̂�𝐢

(a What is the direction of propagation?

(b) What is the wavelength 𝜆𝜆 ?

(c) What is the frequency 𝜈𝜈 ?

(d) What is the amplitude of the magnetic field part of the wave?

(e) Write an expression for the magnetic field part of the wave.




Solution:

(a) From the given electric field vector, we can say that the electric field is directed along the positive 𝑥𝑥 direction. Hence, the direction of motion is along the negative y-direction i.e. −𝚥𝚥̂ .

(b) It is given that,

𝐸𝐸�⃗ = 3.1 N C⁄ cos[(1.8 rad m⁄ ) 𝑦𝑦 + (5.4 × 108 rad s⁄ ) 𝑑𝑑] 𝚤𝚤̂… (1)

The general equation for the electric field vector in the positive 𝑥𝑥 direction can be written as:

𝐸𝐸�⃗ = 𝐸𝐸0 sin(𝑘𝑘𝑥𝑥 − 𝜔𝜔𝑑𝑑)𝚤𝚤̂… (2)

On comparing equations (1) and (2), ‘

we can find

Electric field amplitude, 𝐸𝐸𝑜𝑜 = 3.1 N C⁄

Angular frequency, 𝜔𝜔 = 5.4 × 108 rad s⁄

Wave number, k = 1.8 rad m⁄

The wavelength, 𝜆𝜆 = 2𝜋𝜋1.8

= 3.490 m

(c) Frequency of wave is given as:

𝑣𝑣 =𝜔𝜔2𝜋𝜋

=5.4 × 108

2𝜋𝜋= 8.6 × 107 Hz

(d) Magnetic field strength is given as:


Where 𝑐𝑐 = Speed of light = 3 × 108 m s⁄

∴ 𝐵𝐵0 =3.1

3 × 108= 1.03 × 10−7 T

(e) After seeing the given vector field, it can be observed that the magnetic field vector is directed along the positive 𝑧𝑧 direction. Hence, the general equation for the magnetic field vector is written as:

𝐵𝐵�⃗ = 𝐵𝐵0 cos(𝑘𝑘𝑦𝑦 + 𝜔𝜔𝑑𝑑) 𝑘𝑘�

= {(1.03 × 10−7 T) cos[(1.8 rad m⁄ )𝑦𝑦 + (5.4 × 106 rad s⁄ ) 𝑑𝑑]} 𝑘𝑘�

8.12. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation




(a) at a distance of 1m from the bulb?

(b) at a distance of 10 m? Assume that the radiation is emitted isotopically and neglect reflection.

Solution:

Given that,

The power rating of the bulb, 𝑃𝑃 = 100 W

It is given that about 5% of its power is converted into visible radiation.

Power of visible radiation, 𝑃𝑃′ = 5100

× 100 = 5 W

Hence, the power of visible radiation is 5W.

(a) The distance of a point from the bulb, 𝑑𝑑 = 1 m

Hence, the intensity of radiation at that point is given as:

𝐼𝐼 =𝑃𝑃′

4𝜋𝜋𝑑𝑑2

=5

4𝜋𝜋(𝐼𝐼)2= 0.398 W m2⁄

(b) The distance of a point from the bulb, 𝑑𝑑1 = 10 m

Hence, the intensity of radiation at that point is given as:

𝐼𝐼 =𝑝𝑝′

4𝜋𝜋(𝑑𝑑1)2

=5

4𝜋𝜋(10)2 = 0.00398 W m2⁄

8.13. Use the formula λmT = 0.29 cm K to obtains the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Solution:

At a particular temperature, a body produces a continuous spectrum of wavelengths. In the case of a black body, we can find the wavelength corresponding to the maximum intensity of radiation by Planck's law. It can be given by the relation,

𝜆𝜆𝑚𝑚 = 0.29𝑇𝑇

cm

Where, 𝜆𝜆𝑚𝑚 = maximum wavelength and 𝑇𝑇 = temperature.

Thus, the temperature for different wavelengths can be obtained as:

For 𝜆𝜆𝑚𝑚 = 10−4cm; 𝑇𝑇 = 0.2910−4

= 2900𝑜𝑜 K




For 𝜆𝜆𝑚𝑚 = 5 × 10−5 cm; T = 0.295×10−4

= 5800𝑜𝑜 K

For λm = 10−6cm; T = 0.2910−6

= 290000o K.

The obtained values tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.

8.14. Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.

(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).

(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).

(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].

(d) 5890 Å − 5896 Å [double lines of sodium]

(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].

Solution:

(a) 21 cm come under Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.

(b) Radio waves; it belongs to the short wavelength end

(c) Temperature, 𝑇𝑇 = 2.7oK,

λm is given by Planck's law as:

𝜆𝜆𝑚𝑚 =0.292.7

= 0.11 cm

This wavelength corresponds to microwaves.

(d) wavelength range comes under the yellow light of the visible spectrum.

(e) Transition energy is given by the relation,

𝐸𝐸 = ℎ𝑣𝑣 Where, ℎ = Planck's constant = 6.6 × 10−34 J s

𝑣𝑣 = Frequency of radiation

Energy, 𝐸𝐸 = 14.4 𝑘𝑘eV

∴ 𝑣𝑣 =𝐸𝐸ℎ




=14.4 × 103 × 1.6 × 10−19

6.6 × 10−34

= 3.4 × 1018 Hz

3.4 × 1018 Hz comes under X-rays.

8.15. Answer the following questions:

(a) Long-distance radio broadcasts use short-wave bands. Why?

(b) It is necessary to use satellites for long-distance TV transmission. Why?

(c) Optical and radiotelescopes are built on the ground, but X-ray astronomy is possible only from satellites orbiting the earth. Why?

(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

(f) Some scientists have already predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a bad effect on life on earth. What might be the basis of this prediction?

Solution:

(a) Shortwave frequencies are capable of reaching any location on the Earth because they can be reflected by the ionosphere.

(b) It is necessary to use satellites for long-distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are mainly not reflected by the ionosphere. Hence, satellites are very helpful in reflecting TV signals in a very big area on the earth.

(c) The earth's atmosphere cannot absorb visible light and radio waves. But X-rays being of much smaller wavelength are absorbed by the atmosphere. Therefore, X-ray astronomy is possible only from satellites orbiting the earth at the height of 36000 km above the earth's surface. At such a height, the atmosphere is very thin, and X-rays are not absorbed.

(d) Ozone layer on the top of the atmosphere is very important for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth's surface. In the presence of ultraviolet, it would be very difficult for anything to survive on the surface. Plants cannot live and grow in heavy ultraviolet radiation, nor can the plankton that serves as food for most of the ocean life. The ozone layer acts as a shield to absorb the UV rays, and keep them from doing damage at the Earth's surface.




(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.

(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce smoke that would cover most of the part of the sky, thereby preventing solar light from reaching the atmosphere. Also, it will cause the depletion of the ozone layer.

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cbse ncert solutions for class 12 physics chapter 8...class- xii-cbse-physics electromagnetic waves...

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